Question

Each edge of a cube increased by 10% . Find the
percentage increase in the lateral surface area of the
cube?
(a) 21%
(b) 20%
(c) 19%
(d) Cannot be determined​

Answer 1

Hello sureshagbsc97 !

Answer:

(a) 21 %

Step-by-step explanation:

Given :

Edge of the cube is increased by 10%.

Formulae :

Lateral Surface Area of a Cube = 4a² units²

• Where a is the side of the cube.

Percentage Change = $\frac{F - I}{I} \times 100$ %, Where

• F is the Final Value
• I is the Initial Value

Procedure :

Let the initial side be s.

⇒ LSA = 4s² units² (Initial Value)

s ⇒ s + (10/100)s = (110/100)s

Hence LSA = 4[(110/100)s]²

⇒ LSA = 4s² × 110 × 110 / (100 × 100)

⇒ LSA = 4s² × 121 / 100 units² (Final Value)

Hence Percentage change = $\frac{(4s^{2} \times \frac{121}{100}) - 4s^{2} }{4s^{2}}$ × 100 %

⇒ Percentage change = $\frac{4s^{2} (\frac{121}{100} -1) }{4s^{2}}$ × 100 %

⇒ Percentage change = $\frac{21}{100}$ × 100 %

∴ Percentage change = 21 %

As the Percentage change is positive, the LSA increased by 21 %.

Hence the answer is option (a), 21%.

Thanks !